# Help with finding normal modes of a bar swinging on a string

1. Apr 9, 2013

### BiotFartLaw

1. The problem statement, all variables and given/known data
A bar with mass m and length l is attached at one end to a string (also length l) and is swinging back and forth. Find the normal modes of oscillation.

2. Relevant equations
L=T-U, and the Lagrange-euler equation, I=(1/12)ml^2

3. The attempt at a solution
So my idea is this. Use the two angles $\theta$ (angle of string from the vertical) and $\phi$ (angle of the bar from the vertical) as the generalized coordinates and set up a Lagrangian to get two (probably coupled) differential equations. THen I'm guessing two different eigenfrequencies will pop out when I solve them.

My problem is that I'm not sure what the (translational) kinetic energy is. (The rotational would just be I*ω^2 and this ω^2 would, in the end, give me the two frequencies [?]) I've tried finding $\dot{x}$ and $dot{y}$ using sines and cosines of the angles, but when I do (and I make appropriate approximations) I'm left with no $dot{\theta}$ or $dot{\phi}$terms which doesn't seem right.
But I'm also not sure if it's as simple as $T=ml^2(dot{\theta}+dot{phi})$.

Thanks

Last edited: Apr 9, 2013
2. Apr 9, 2013

### vela

Staff Emeritus
The easiest thing to do is simply write down the x and y coordinates of the center of mass of the rod in terms of the two angles and then calculate $\frac{1}{2}(\dot{x}^2+\dot{y}^2)$.

3. Apr 9, 2013

### BiotFartLaw

Thanks. A new question though:

What is ω (from the rotational KE) in terms of the two angles? Usually $\omega = dot{\theta}$. But that's only for one angle. Because if ω isn't in terms of the angle(s) then the rotational KE drops out when you take the derivatives for L which doesn't seem right.

Thanks again.

4. Apr 9, 2013

### vela

Staff Emeritus
From the way you defined the angles, you should have $\omega=\dot{\phi}$, right? Changing that angle corresponds to the bar rotating whereas changing $\theta$ only causes translation of the bar.